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If a, b, c are positive real numbers, then `a^("log"b-"log"c) xx b^("log"c-"log"a) xx c^("log"a - "log"b)` |
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Answer» Correct Answer - B We have, `"log" {a^("log"b-"log"c) xx b^("log"c - "log" a) xx c^("log"a-"log"b)}` `= ("log" b-"log"c)"log" a + ("log"c-"log"a)"log"b + ("log"a-"log"b)"log"c ` = 0 `a^("log"b-"log"c) xx b^("log"c - "log" a) xx c^("log"a-"log"b)=1` |
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